Problem: $ \left(\dfrac{4}{9}\right)^{-\frac{3}{2}}$
$= \left(\dfrac{9}{4}\right)^{\frac{3}{2}}$ $= \left(\left(\dfrac{9}{4}\right)^{\frac{1}{2}}\right)^{3}$ To simplify $\left(\dfrac{9}{4}\right)^{\frac{1}{2}}$ , figure out what goes in the blank: $\left(? \right)^{2}=\dfrac{9}{4}$ To simplify $\left(\dfrac{9}{4}\right)^{\frac{1}{2}}$ , figure out what goes in the blank: $\left({\dfrac{3}{2}}\right)^{2}=\dfrac{9}{4}$ so $ \left(\dfrac{9}{4}\right)^{\frac{1}{2}}=\dfrac{3}{2}$ So $\left(\dfrac{9}{4}\right)^{\frac{3}{2}}=\left(\left(\dfrac{9}{4}\right)^{\frac{1}{2}}\right)^{3}=\left(\dfrac{3}{2}\right)^{3}$ $= \left(\dfrac{3}{2}\right)^{3}$ $= \left(\dfrac{3}{2}\right)\cdot\left(\dfrac{3}{2}\right)\cdot \left(\dfrac{3}{2}\right)$ $= \dfrac{9}{4}\cdot\left(\dfrac{3}{2}\right)$ $= \dfrac{27}{8}$